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STRUCTURE OF Ca-40 AND Ca-48
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPT . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Structure of magic nucleus Ca-40 with S = 0 In Fig. 7a of my published paper removing the extra neutrons one can see the observable sides of Mg-24 with the two alpha particles. In fact the Ca-40 consists of the Mg-24 when we add 4α particles which give the symmetrical shape with equal 20 protons and 20 neutrons having a total spin S = 0. Such a symmetrical shape belongs to the magic nuclei. In fig. 7a you can see also that at some positions there exist not only four or five bonds per nucleon but six ones. Therefore Ca-40 contains lattice points as many as possible with the greatest coordination number of 6, which contributes to the increase of the binding energies of pn or np bonds. In the following diagram one observes the structure of Mg-24 from p1 to n12 with S = 0. Here we also observe the two alpha particles of p13, n13, p14, n14, p15, n15, p16, and n16 with S = 0. On purpose the other two alpha particles with p17, n17, p18, n18, p19, n19, p20, and n20 (in front of the Mg-24 and behind it, with S = 0 are not shown . Therefore Ca-40 consists of 20p and 20n. Here one concludes also that the great number of pp repulsions of long range cannot overcome the pn bonds because in some positions there exist six bonds per nucleon which contribute to the increase of the binding energies. Structure of magic nucleus Ca-48 with S = 0 In Fig 7a of my published paper one sees that the alpha particles along with the shape of Mg-24 make blank positions able to receive extra neutrons. In fact, the blank positions are able to receive 8 extra neutrons which form two np bonds per neutron (non single np bonds). In Fig. 7a of my published paper one sees also four extra neutrons indicated by small circles. In the following diagram you can see only the two extra neutrons like n17, and n18 with positive spins because they belong to the second horizontal plane. You can see also the other two extra neutrons like n19 and n20 with negative spins because they belong to the fifth horizontal plane. Each of such extra neutrons makes two np bonds per neutron in order to contribute more to the increase of the binding energies. For example at point n20 we observe the two np bonds per extra neutron as (n20-p10) and (n20-p16). On purpose the other four extra neutrons like n21, n22, n23 and n24 are not shown here.' ' ' ' ' STRUCTURE OF Ca-48' ' (The structure consists of 6 horizontal planes existing from the +HP1 with positive spins to -HP6 with negative spins.The p17, n17, p18, n18, p19, n19, p20, n20 of the two α particles in front of Mg-24 and behind it, as well as the extra neutrons n21, n22, n23, and n24 are not shown here)' ' p12..........n12' ' n11..........p11 -HP6' ' n10..........p10..........n20' ' n19..........p9............n9 +HP5' ' n14..........p8............n8............p16' ' p14..........n7............p7...........n16 -HP4' ' p13..........n6............p6............n15' ' n13..........p5...........n5............p15 +HP3' ' n17.........p4............n4' ' n3............p3............n18 -HP2' ' n2............p2' ' p1...........n1 +HP1' ' ' Category:Fundamental physics concepts